A lot of variables

Let, $a^{x}=b^{y}=c^{z}=k$ Then $a=k^{\frac{1}{x}} , b=k^{\frac{1}{y}} , c=k^{\frac{1}{z}}$ Put the value of $a$ , $b$ , $c$ in the given relation $a^{3}=b^{2}c$ we get, $k^{\frac{3}{x}}=k^{\frac{2}{y}}k^{\frac{1}{z}}$ or, $k^{(\frac{3}{x}-\frac{2}{y})}=k^{\frac{1}{z}}$ hence $(\frac{3}{x}-\frac{2}{y})=\frac{1}{z}$

If we assume that $a,b,c$ are positive reals, then if we take the common logs of the given equations we find that

$x\log(a) = y\log(b) = z\log(c)$ and $3\log(a) = 2\log(b) + \log(c).$

From the first of these we see that $\log(c) = \dfrac{y}{z}\log(b),$ and so if we also assume that none of $a,b,c$ equal $1$ we find that

$\dfrac{3\log(a)}{x\log(a)} = \dfrac{2\log(b) + \dfrac{y}{z}\log(b)}{y\log(b)} \Longrightarrow \dfrac{3}{x} = \dfrac{2 + \dfrac{y}{z}}{y} = \dfrac{2}{y} + \dfrac{1}{z} \Longrightarrow \dfrac{3}{x} - \dfrac{2}{y} = \boxed{\dfrac{1}{z}}.$

If a^x=b^y=c^z, then we can say that a^x a^x=b^y c^z a^2x=b^y c^z Compare this to a^3=b^2 c Hence, 2x=3, y=2, and z=1 x=3/2 Substitute these values into the equation into (3/x-2/y) If we do this, we arrive at the answer 2-1=1. Now we need to choose the answer choice that will get us to 1. Since z=1, 1/z=1 as well, so the answer is 1/z.